How to display lines 2-4 after each grep result?

Question

I'm parsing a mailbox file that stores e-mail server reports for unsuccessfully delivered e-mail. I wish to extract bad e-mail addresses, so that I remove them from the system. The log file looks like this:

...some content...
                   The mail system

<[email protected]>: host mx1.hotmail.com[65.54.188.94] said: 550
    Requested action not taken: mailbox unavailable (in reply to RCPT TO
    command)

...some content...
                   The mail system

<[email protected]>: host viking.optimumpro.net[79.101.51.82] said: 550
    Unknown user (in reply to RCPT TO command)

...some content...
                   The mail system

<[email protected]>: host mta5.am0.yahoodns.net[74.6.140.64] said: 554
    delivery error: dd This user doesn't have a yahoo.com account
    ([email protected]) [0] - mta1172.mail.sk1.yahoo.com (in reply to end
    of DATA command)

...etc.

E-mail address comes 2 lines after a line with "The mail system". Using grep like this gives me the "The mail system" line and the next two lines:

grep -A 2 "The mail system" mbox_file

However, I don't know how to remove the "The mail system" line and the second empty line from this output. I guess I could write PHP/Perl/Python script to do it, but I wonder if this is possible with grep or some other standard tool. I tried to give negative offset to -B parameter:

grep -A 2 -B -2 "The mail system" mbox_file

But grep complains:

grep: -2: invalid context length argument

Is there a way to do this with grep?

Answer 1

The simplest way to solve it using grep only, is to pipe one more inverted grep at the end. For example:

grep -A 4 "The mail system" temp.txt | grep -v "The mail system" | grep -v '^\d*$'

Answer 2

If you aren't locked in to using grep, try sed ...

sed -n '/The mail system/{n;n;p}' 

When it finds a line containing "The mail system", it reads the next line twice, via the n;n;, discarding each previous line as it does so.
This leaves the 3rd line of your group in the pattern space, which is then printed via sed's p command.. The leading -n option prevents all other printing.

To print the next two lines as well, it is just a case of next and print n;p twice more.

sed -n '/The mail system/{n; n;p; n;p; n;p}'   

The next-line reads for the lines you require can be accumulated and printed a a single block with just one p... N reads the next line and appends it to the pattern space,

Here is the final condensed version...

sed -n '/The mail system/{n;n;N;N;p}'   

If you want a group seperator, similar to what grep wouuld output, you can use sed's insert command i (which must be the last command on a line)...

Here is the syntax to include a group seperator

sed -n '/The mail system/{n;n;N;N;p;i--
       }' > output-file  # or | ...

Here is the output for the first match:

<[email protected]>: host mx1.hotmail.com[65.54.188.94] said: 550
    Requested action not taken: mailbox unavailable (in reply to RCPT TO
    command)                                                                    
--

Answer 3

grep -A 2 -B -2 "The mail system" mbox_file

-B is for previous lines, so no need to give -negative value.

grep -A 2 -B 2 "The mail system" mbox_file   # This will work please check

Answer 4

I see no point in using only grep(s), except if that's a strict constraint. It cannot be done with one call to grep.

grep -A 2 "The mail system" mbox_file | tail -n +3

• grep: Find the line and output 2 lines after,
• tail: cut the first 2 lines (i.e. start from the third line).

Answer 5

If yo want to remove the first 2 lines pipe it to sed

sed '1,2d'

as in

grep -A 2 "The mail system" mbox_file | sed '1,2d'

Answer 6

This prints the next 1 line following the regexp match, using Perl

perl -ne 'print if( (/The mail system/ && ($end=1))..!$end-- )'